Carbon-14 is a radioactive isotope which decays with a half-life of 5730 years. what is the first-order rate constant for its decay, in units of years–1?

In chemical equilibrium, radioactive decay occurs because of unstability due to the high neutron-to-proton ratio. Through time, the radioactive element is converted to a new element. This radioactive decay undergoes first order reaction. Its equation is in the form of 

A = A₀e^(-kt), where A is the amount of the element after time t, A₀ is the original amount of element at t=0 and k is the rate constant. 

Half-life is the amount of time for A to be 1/2 of A₀. Modifying the equation during half time, we let A=1/2 A₀. Then.

1/2 A₀ = A₀e^(-kt)
1/2 = e^(-k(5730))
k = 0.000121

Thus, the first-order rate constant is equal to 0.000121 per year.

So what you would do is go through and add 3 teaspoons(tsp) 6 times because of the 6 tablespoons(tbsp). which will give you an answer of 18. another way is to multiply 3x6=18.

It could possibly be beneficial if the experiment is successful.